And conclude the limit is indeed $0$ through any line. (a formal justification involves epsilon-delta, but I omit it here because that is another question for another time).

I am thinking that the first equality sign is wrong.

Remark Most books I've read seem to do everything without the limit operator. Stewart for instance just argues the limit is this and this along this path and that path. I want to do my answers with the limit operators

You’re right: the first equality is premature. You’ve shown only that the limit is $0$ when you approach the origin along a straight line. To claim that the limit is actually $0$, you have to show that it’s $0$ no matter how you approach the origin; there are functions that behave very nicely along straight lines but not along more complicated paths.
–
Brian M. ScottFeb 6 '13 at 2:08

You're right. The expression in the middle does not make sense.
–
1015Feb 6 '13 at 2:09

The second equality sign is justified though right? Provided I made my argument that "we are going to take the limit along this and that path" and then my conclusion should be followed cleanly without any loss in rigor or error?
–
HawkFeb 6 '13 at 2:10

For the second and third terms to make sense, you need to replace $(x,y)\rightarrow(0,0)$ by $x\rightarrow 0$.
–
1015Feb 6 '13 at 2:11

2 Answers
2

There are so many ways to approach (0; 0), appoaching along the lines $y = mx$ is 1 way, other ways are approaching along the curves $y = x^2$, or $y = \sqrt{x}$, or $y = x^3$, or... any curves that pass through (0; 0).

So, pointing out that as $(x; y) \rightarrow (0; 0)$ along the lines $y = mx$ the limit is 0, is definitely not enough to show that the limit does exist.

You can change it into polar co-ordinate, like this:
Let $\left\{ \begin{array}{l} x = r \cos \varphi \\ y = r \sin \varphi \end{array} \right.$, when $(x; y) \rightarrow (0; 0)$ it means that $r \rightarrow 0$, and $\varphi$ can vary freely. So, to show that the limit exists, all you must do is to show that as $r \rightarrow 0$, and $\varphi$ takes any value, the limit stays the same.

Just to see how bad things can be while being totally nice along straight lines. Consider the function $f:\mathbb R^2 \to \mathbb R$ that is defined like this. For any point $(x,y)$ consider the slope of the line through the origin on which it lies (take care of the case $(x,y)=(0,0)$ separatel). Then if the slope is not of the form $1/n$, $n\in \mathbb N$ then set $f(x,y)=0$. Otherwise $(x,y)$ lies on a line through the origin of slope $1/n$ and then set $f(x,y)=0$ if the distance from $(x,y)$ to the origin is less then $1/n$, and set $f(x,y)=1$ otherwise. The limit at the origin along any straight line is $0$ though the global limit at the origin does not exist.